3.245 \(\int (\frac {c}{a+b x^2})^{3/2} \, dx\)

Optimal. Leaf size=21 \[ \frac {c x \sqrt {\frac {c}{a+b x^2}}}{a} \]

[Out]

c*x*(c/(b*x^2+a))^(1/2)/a

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6720, 191} \[ \frac {c x \sqrt {\frac {c}{a+b x^2}}}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2),x]

[Out]

(c*x*Sqrt[c/(a + b*x^2)])/a

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx &=\left (c \sqrt {\frac {c}{a+b x^2}} \sqrt {a+b x^2}\right ) \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx\\ &=\frac {c x \sqrt {\frac {c}{a+b x^2}}}{a}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \[ \frac {c x \sqrt {\frac {c}{a+b x^2}}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x^2))^(3/2),x]

[Out]

(c*x*Sqrt[c/(a + b*x^2)])/a

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fricas [A]  time = 0.46, size = 19, normalized size = 0.90 \[ \frac {c x \sqrt {\frac {c}{b x^{2} + a}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c/(b*x^2+a))^(3/2),x, algorithm="fricas")

[Out]

c*x*sqrt(c/(b*x^2 + a))/a

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giac [A]  time = 0.48, size = 28, normalized size = 1.33 \[ \frac {c^{2} x \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c x^{2} + a c} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c/(b*x^2+a))^(3/2),x, algorithm="giac")

[Out]

c^2*x*sgn(b*x^2 + a)/(sqrt(b*c*x^2 + a*c)*a)

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maple [A]  time = 0.00, size = 26, normalized size = 1.24 \[ \frac {\left (b \,x^{2}+a \right ) \left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}} x}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/(b*x^2+a)*c)^(3/2),x)

[Out]

(b*x^2+a)/a*x*(1/(b*x^2+a)*c)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {c}{b x^{2} + a}\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c/(b*x^2+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((c/(b*x^2 + a))^(3/2), x)

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mupad [B]  time = 2.74, size = 19, normalized size = 0.90 \[ \frac {c\,x\,\sqrt {\frac {c}{b\,x^2+a}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x^2))^(3/2),x)

[Out]

(c*x*(c/(a + b*x^2))^(1/2))/a

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sympy [A]  time = 1.46, size = 66, normalized size = 3.14 \[ \begin {cases} c^{\frac {3}{2}} x \left (\frac {1}{a + b x^{2}}\right )^{\frac {3}{2}} + \frac {b c^{\frac {3}{2}} x^{3} \left (\frac {1}{a + b x^{2}}\right )^{\frac {3}{2}}}{a} & \text {for}\: a \neq 0 \\- \frac {c^{\frac {3}{2}} x \left (\frac {1}{b}\right )^{\frac {3}{2}} \left (\frac {1}{x^{2}}\right )^{\frac {3}{2}}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c/(b*x**2+a))**(3/2),x)

[Out]

Piecewise((c**(3/2)*x*(1/(a + b*x**2))**(3/2) + b*c**(3/2)*x**3*(1/(a + b*x**2))**(3/2)/a, Ne(a, 0)), (-c**(3/
2)*x*(1/b)**(3/2)*(x**(-2))**(3/2)/2, True))

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